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I have a standard USB adapter of 5 volts / 1 Amp and I would like to instead of cutting off voltage with resistor, I need to limit the maximum output to 200mA. I'm confused about how to accomplish this, because as far as I understood, if I would use a 6.8ohm resistor, it would cut down the voltage to 3.7v and cut down some amount of current, but the maximum potential of the current that can flow into my circuit will be still more than 200mA, correct?

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  • \$\begingroup\$ What is "a standard USB adapter of 5v/1mA"? \$\endgroup\$ – Bort Nov 8 '16 at 0:29
  • \$\begingroup\$ @bort I think he meant 5v and 1Amp \$\endgroup\$ – Allahjane Nov 8 '16 at 0:33
  • \$\begingroup\$ @Bort sorry I meant 1amp \$\endgroup\$ – 0x29a Nov 8 '16 at 0:40
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If you assumed your load was linear ( unlikely) then a 200mA drop across 6.8 Ohms= 1.26V drop from 5V= 3.74V

However if your load did not consume 200mA the voltage would rise. and if it consumed more the voltage drops.

Thus the solution is an LDO regulator which has a a dropout<= 1V at the desired voltage and current limit and temperature rise from your heatsink.

CHeapest solution is http://www.digikey.com/product-detail/en/microchip-technology/MIC94325YMT-TR/576-4164-1-ND/3681264 with 0.1V dropout at 0.5A so it you wanted to limit current, then choose ~6 Ohms

Or use a Thruhole part http://www.digikey.com/product-detail/en/texas-instruments/TPS7201QP/296-8052-5-ND/373257

Both are adjustable LDO's with R ratio to choose outout voltage.

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  • \$\begingroup\$ Thanks! My load is a battery and I'm trying to figure out IF I can use a resistor to prevent too much current going to it. \$\endgroup\$ – 0x29a Nov 8 '16 at 0:44
  • \$\begingroup\$ no, you can use a 3.7V LDO with a series R \$\endgroup\$ – Tony Stewart EE75 Nov 8 '16 at 0:53
  • \$\begingroup\$ @0x29a - "My load is a battery" - Now that you have revealed this: Are you sure that you want to charge whatever battery you have, with just a 3.7V voltage source? If the battery is Li-Ion or Li-Po (which, at a guess, seems likely for the voltage you were trying to generate), then you cannot safely charge it like that. \$\endgroup\$ – SamGibson Nov 8 '16 at 10:34
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Assuming you meant 5v and 1 ampere , instead of 5v and 1mA at which getting a 200mA out of it at 5v would be impossible.

You can simply put a limiting resistor of 5/0.2 = 25ohm in series with the load but this is only valid if there is no voltage drop across your load.

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  • \$\begingroup\$ Thanks! I meant 1amp, but the output needs to be 3.7v 200mA not 5v 200mA :) \$\endgroup\$ – 0x29a Nov 8 '16 at 0:39

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